Decompress File Paths in Java

A comma separated list of strings can be compressed by finding common substrings and grouping those strings together into Token Lists

More formally, a Token List is a list where each item can be one of the following:

a string
a concatenation of Token Lists

Examples:

ab,ac => a{b,c}
ac,ad,bc,bd => {a,b}{c,d}
axc,axd,bxc,bxd => {a,b}x{c,d}

This technique is used to compress folder paths since a folder like

README.md
resources/
    js/
        app.js
        boot.js
        comp/
            a.js
            b.js

can be represented as a list of strings:

[
    "README.md",
    "resources/js/app.js",
    "resources/js/boot.js",
    "resources/js/comp/a.js",
    "resources/js/comp/b.js"
]

and further encoded into comma separated string

“README.md,resources/js/app.js,resources/js/boot.js,resources/js/comp/a.js,resources/js/comp/b.js”

and then we can compress them to the following Token List:

README.md,resources/js/{app,boot,comp/{a,b}}.js

Given a Token List representing a compressed folder, return the initial list of paths.

Example 1:

Input:
"resources/{js/{app.js,comp/{a.js,b.js},boot.js}},README.md"

Output: [
    "README.md",
    "resources/js/app.js",
    "resources/js/boot.js",
    "resources/js/comp/a.js",
    "resources/js/comp/b.js"
]

Example 2:

Input:
"my_{first,second}_folder/{a,b}{.js,.py}"

Output: [
    "my_first_folder/a.js,
    "my_first_folder/a.py,
    "my_first_folder/b.js,
    "my_first_folder/b.py,
    "my_second_folder/a.js,
    "my_second_folder/a.py,
    "my_second_folder/b.js,
    "my_second_folder/b.py
]

Note: The resulting array can be in any order

Understanding the Problem

The core challenge of this problem is to expand a compressed representation of file paths into their full list of paths. This is significant in applications where file paths are stored in a compressed format to save space, and we need to retrieve the original paths for processing.

Potential pitfalls include correctly handling nested token lists and ensuring all combinations are generated.

Approach

To solve this problem, we need to recursively expand the token lists. Here's a step-by-step approach:

Identify the base case: If the input string does not contain any braces, it is a single path.
For strings with braces, find the first pair of braces and extract the content within.
Split the content within the braces by commas to get the individual options.
Recursively expand the rest of the string and concatenate each option with the expanded rest of the string.
Combine all results and return.

Algorithm

Here's a detailed breakdown of the algorithm:

Initialize a list to store the results.
Find the first pair of braces in the string.
Extract the prefix before the braces, the content within the braces, and the suffix after the braces.
Split the content within the braces by commas to get the individual options.
Recursively expand the suffix.
For each option, concatenate it with the prefix and each expanded suffix, and add the result to the list.
Return the list of results.

Code Implementation

import java.util.ArrayList;
import java.util.List;

public class DecompressFilePaths {
    public static List<String> decompress(String compressed) {
        List<String> result = new ArrayList<>();
        decompressHelper(compressed, result, "");
        return result;
    }

    private static void decompressHelper(String compressed, List<String> result, String current) {
        int openBrace = compressed.indexOf('{');
        if (openBrace == -1) {
            // Base case: no more braces, add the current path to the result
            result.add(current + compressed);
            return;
        }

        int closeBrace = findMatchingBrace(compressed, openBrace);
        String prefix = compressed.substring(0, openBrace);
        String suffix = compressed.substring(closeBrace + 1);
        String[] options = compressed.substring(openBrace + 1, closeBrace).split(",");

        for (String option : options) {
            decompressHelper(suffix, result, current + prefix + option);
        }
    }

    private static int findMatchingBrace(String str, int openBrace) {
        int balance = 0;
        for (int i = openBrace; i < str.length(); i++) {
            if (str.charAt(i) == '{') balance++;
            if (str.charAt(i) == '}') balance--;
            if (balance == 0) return i;
        }
        throw new IllegalArgumentException("Unmatched braces in the input string");
    }

    public static void main(String[] args) {
        String input1 = "resources/{js/{app.js,comp/{a.js,b.js},boot.js}},README.md";
        System.out.println(decompress(input1));

        String input2 = "my_{first,second}_folder/{a,b}{.js,.py}";
        System.out.println(decompress(input2));
    }
}

Complexity Analysis

The time complexity of this approach is O(n * m), where n is the number of options within the braces and m is the length of the string. The space complexity is O(n) due to the recursion stack and the storage of results.

Edge Cases

Potential edge cases include:

Empty input string: Should return an empty list.
Unmatched braces: Should throw an exception.
Nested braces: Should handle correctly.

Examples of edge cases and their expected outputs:

Input: ""
Output: []

Input: "a{b,c{d,e}}f"
Output: ["abf", "acdf", "acef"]

Testing

To test the solution comprehensively, include a variety of test cases:

Simple cases with no braces.
Cases with single-level braces.
Cases with nested braces.
Edge cases as discussed above.

Use a testing framework like JUnit to automate the tests.

Thinking and Problem-Solving Tips

When approaching such problems:

Break down the problem into smaller parts.
Identify the base case and recursive case.
Use helper functions to keep the code clean and modular.
Test with simple cases first before moving to complex ones.

Practice similar problems to improve problem-solving skills.

Conclusion

In this blog post, we discussed how to decompress file paths represented as token lists. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for efficient data processing and storage.

Keep practicing and exploring further to enhance your skills.

Additional Resources

For further reading and practice problems, check out:

Decompress File Paths in Java with Time Complexity Analysis